$A$ student measured the length of a rod and wrote it as $3.50\;cm$. Which instrument did he use to measure it?

Which of the following is the most precise device for measuring length:
$A)$ a vernier callipers with $20$ divisions on the sliding scale
$B)$ a screw gauge of pitch $1 \; mm$ and $100$ divisions on the circular scale
$C)$ an optical instrument that can measure length to within a wavelength of light?

In a Vernier calipers, when both jaws touch each other, the zero of the Vernier scale is shifted to the right of the zero of the main scale and the $7^{th}$ Vernier division coincides with a main scale division. If the value of $1$ main scale division is $1 \text{ mm}$ and there are $10$ Vernier scale divisions, then the Vernier caliper has

In a Vernier Calipers,$10$ divisions of the Vernier scale are equal to $9$ divisions of the main scale. When both jaws of the Vernier calipers touch each other,the zero of the Vernier scale is shifted to the left of the zero of the main scale,and the $4^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. One main scale division is equal to $1\,mm$. While measuring the diameter of a spherical body,the body is held between the two jaws. It is observed that the zero of the Vernier scale lies between $30$ and $31$ divisions of the main scale,and the $6^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. The diameter of the spherical body is $.......\,cm$.

$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.